It's been a long time since I posted here.
Here is a puzzle that can be solved by 9th graders, but can cause some confusion with some people. Assuming a triangle is defined by 6 characteristics- 3 lengths of sides and 3 angles, can there be 2 triangles that have 5 of the 6 categories equal to one another, but still not be identical triangles? If so, find those triangles. If not, prove it cant be so.

Have fun,

Nir

If you're just defining a triangle as 3 lengths of sides and 3 angles, then I can draw you a million such triangles as you suggest.

If instead, you define a triangle as a closed figure with three sides, and three internal angles that sum to 180 degrees, then the puzzle becomes a little trickier.

Scott - 3 internal angles and a euclidian surface - all angles sum to 180 degrees. No tricks.

For some reason I'm getting stuck on the ASA and SAS proofs of similarity. Taking them as ASASAS or SASASA, working around one triangle, I don't see how losing one of the letters can give you a non congurent triangle. Then again I am at work at the moment.

Jonathan,

First, thanks for clarifying the terminology - English is not my native language. But you have some fallacy in your "proof". Think again.

Nir

Imagine two triangles: triangle ABC with opposite sides a, b, and c; and triangle A'B'C', with opposite sides a', b', and c'. Now imagine angle A = A', B=B', and C=C'. Also imagine side AC = side B'C', and side BC = side A'B'. Thus there are five corresponding equalities. Yet these two triangles are not necessarily identical.

A bit hard to explain without a diagram, but draw the above and you'll see what I mean.

Jack Shalom

Jack,

There is a "family" of such triangles that will fullfill these conditions. Can you find them?

Nir

Here's a step towards defining a family: imagine two right triangles. The first triangle has sides a, b, and hypotenuse c; the second triangle has sides b, c, and hypotenuse d. If these two triangles are similar, they will fulfill the requirements of the problem. The triangles are similar if b/a = c/b. Then, using Pythagorean theorem, you can conclude that the two triangles are similar when b exponent 4 equals a exponent 4 + a exponent 2 * b exponent 2. Whether this has real solutions, I'm not sure.

If there are real solutions to the above, then you could probably extend this to non-right triangles using the Law of Cosines, rather than the Pythagorean theorem to set up the needed relationships amongst the sides.

Am I getting close?

Jack

Because 5 characteristics have to be equal, either all the angles are the same or all the sides are the same. If the sides were the same, they would be the same triangle (ie 6 would match). That means all the angles are the same. Therefore, the triangles will be similair triangles (ie one of the triangles is a scaled up/down version of the other).

Consider a triangle with sides of lengths x,y and z. (Starting with smallest). Because they're identical triangles, the other will have lengths of kx, ky, and kz, where k is a constant. To fit the boundaries of Nir's puzzle, kx=y and ky=z.

In triangle x,y,z
kx=y ky=z and x+y>z
z=k sqr times x
x+kx>ksqr x
1+k>k squared or 1>k(k-1)
Therefore k is between 1 and 2?
K being greater than one, because if it was a fraction, then the second triangle would be smaller. If the second was smaller, we would swap triangles and k would be greater than 1 (i.e. k =1/2 it's same as saying k=2 for the other triangle).
I've simply used 2 via substitution, there's a more exact number if you want to find it by solving 1=k(k-1).

Let's try 1.5.
a 4,6,9 triangle and a 6,9,13.5. Fit the credentials.
And no pythag. or trig in sight.

George

edit:k is less than 1.618 . Therefore answer in full is any triangle with side x,kx and k sqr x and another triangle kx,ksqr x and k cubed x. will work for values where 1<k<1.618.

P.S. how do I do a squared sign on here?

Let me get through the algebra to the kernal here...

If I read you right... the idea is that the triangles are similar...
and share measures in two of their sides which are NOT corresponding.

Going back to the simple notation introduced above, angles a, b and c are identical while sides BC and CA are identical to sides C'A' and A'B'. Dies this work for you Nir?

Jon,

That's the idea--they share two non-corresponding sides. George, in addition, has neatly determined some additional constraints on the lengths of the sides.

Jack

If I understand the question correctly, and the ideas given above: Choose the sides to be x, x/k and x/k^2 where x>0, k>0 and of course k<>1. Construct the other triangle to be kx, x and x/k.

/Tomas

Tomas you got it right...

Hey, I got it right too.
My answer works and also includes the parameters for k. Tomas's answer is just a different way of saying the same thing (see edit in second post)

I have a Word document of such odd characters, some of which I know how to create (by Ctrl + Alt + whatever), and the rest of which I simply copy'n'paste.

There are superfixes in there, but when I copy'n'paste them into a post reply window here, they lose their superfix property.

(And I don't know how to create them.)

Dave

k^2 = "k squared"